Energy associated with electric feild.

[vrinda mukund]

Homework Statement

a conducting sphere of radius R has charge +Q on its surface. if the charge on the sphere is doubled and radius is halved, the energy associated with the electric field will be?

Homework Equations

electric field = K * Q/(R*R)

Energy associated with field will be (1/2)Integral of (ebsilon)E*E

The Attempt at a Solution

what am getting as answer is 64.but that is wrong. how can i solve this?

 

[G01]

Please show your work. See point 1:

https://www.physicsforums.com/showthread.php?t=94380

 

[vrinda mukund]

the answer should be 8 times. E = q/4pi*ebsilon*r*r. if we take it as E1 and E2 in the two cases and then compare them, we get the result. but my doubt is, in the question they have asked to find the energy associated with the electromagnetic field which i think is equal to integral(1/2)ebsilon*E*E. but that won't give the answer. the formula that gives the correct result is actually for finding electric field at a point, rite? how can these two be the same?

 

[G01]

The answer is not 8.

If the radius is halved and the charge is doubled, the electric field is increased by a factor of 4, not 64.

Like you said: The energy depends on the square of the electric field. Thus, if the electric field increases by a factor of 4, by how much does the square of the electric field (and thus the energy) increase by?

 

[vrinda mukund]

but sir, how can it be 4? as Q is double a factor 2 appears on the numerator. again as there is an (r/2) square on the denominator a factor 4 also appears in the numerator. so field in the 2nd case is 4*2=8 times the first one, rite?

 

[G01]

Oops! I forgot to square the 1/2! These little mistakes happen. My apologies for the confusion!

Your answer of 64 is correct. The field is increased by a factor of 8, which means the energy is increased by a factor of 64.

 

[vrinda mukund]

but sir, the thing is that options given does not include such an answer. given options are
1. increases 4 times.
2.increases 8 times
3.remains the same
4.decreases 4 time

the answer should be 'increases 8 times'

 

[G01]

OK Assuming I am not misreading the question. Your answer of 64 is correct, regardless of what the book says. (It may be a typo.)

We agree that the field is increased by a factor of 8:

$$E\propto\frac{Q}{r^2}$$

so, $$E' \propto \frac{2Q}{(1/2)^2r^2}=\frac{8Q}{r^2}=8E$$

Therefore the energy:

$$U\propto E^2$$

and $$U'\propto E'^2=8^2 E^2 = 64 U$$

 

[GrantB]

Looks like a typo of some sort to me. Maybe they meant to ask how much the electric field increases by.